Find $\lim_{h\to 0}\dfrac{7e^{{4+h}}-7e^4}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $e^4$ (Choice B) B $7e^4$ (Choice C) C $7$ (Choice D) D The limit doesn't exist
The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $7e^{{4+h}}-7e^4$, we can tell that the function is $f(x)=7e^x$ and the $x$ -value is $4$. In other words, the limit expression is equal to $f'(4)$ for $f(x)=7e^x$. Let's find $f'(x)$ : $f'(x)=7\cdot e^x=7e^x$ Now let's evaluate $f'(4)$ : $f'(4)=7e^4$ In conclusion, $\lim_{h\to 0}\dfrac{7e^{{4+h}}-7e^4}{h}=7e^4$.